Question

Four metallic plates each with a surface area of one side A,
are placed at a distance d from each other. The two outer
plates are connected to one point A and the two other inner
plates to another point B as shown in the figure. Then the
capacitance of the system is
(a) ℰ₀A/d
(b) 2ℰ₀A/d
(c) 3ℰ₀A/d
(d) 4ℰ₀A/d

Answer 1

Given:

Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B.

To find:

Net capacitance of the system

Calculation:

4 plates are attached in such a way that there will be formation of three capacitors each of them having a potential difference of AB.

Since they have the same potential difference we can considered that they are in parallel combination.

Refer to the attached photo :

Capacitance of each capacitor be C ;

$\boxed{ \sf{C = \dfrac{\epsilon_{0}A}{D} }}$

Now , three search capacitors arranged in parallel combination :

$\sf{C_{net} = \dfrac{\epsilon_{0}A}{D} + \dfrac{\epsilon_{0}A}{D} + \dfrac{\epsilon_{0}A}{D}}$

$\sf{ = > C_{net} = \dfrac{3\epsilon_{0}A}{D} }$

So , final answer is :

$\boxed{ \blue{ \rm{ C_{net} = \dfrac{3\epsilon_{0}A}{D} }}}$